Simplify the following expression and state the condition under which the simplification is valid. You can assume that $a \neq 0$. $r = \dfrac{8}{3a^2 + 10a} \div \dfrac{9}{6(3a + 10)} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{8}{3a^2 + 10a} \times \dfrac{6(3a + 10)}{9} $ When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ 8 \times 6(3a + 10) } { (3a^2 + 10a) \times 9 } $ $ r = \dfrac {8 \times 6(3a + 10)} {9 \times a(3a + 10)} $ $ r = \dfrac{48(3a + 10)}{9a(3a + 10)} $ We can cancel the $3a + 10$ so long as $3a + 10 \neq 0$ Therefore $a \neq -\dfrac{10}{3}$ $r = \dfrac{48 \cancel{(3a + 10})}{9a \cancel{(3a + 10)}} = \dfrac{48}{9a} = \dfrac{16}{3a} $